Answer
$0.0018N*m$
Work Step by Step
We know that
$\alpha=\frac{\omega^2-\omega_{\circ}^2}{2\Delta \theta}$
$\alpha=\frac{(450rev/min \times 2\pi rad/rev\times 1min/60s)^2-(0)^2}{2(3.0rev\times 2rad/rev)}=59rad/s^2$
Now $I=\frac{1}{2}(0.017)(0.060)^2=3.1\times 10^{-5}Kg*m^2$
The torque can be determined as
$\tau=I\alpha$
We plug in the known values to obtain:
$\tau=(3.1\times 10^{-5})(59rad/s^2)=0.0018N*m$
